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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Thu, 5 Sep 2024 12:29:45 +0200
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Op 04.sep.2024 om 15:04 schreef olcott:
> On 9/4/2024 4:34 AM, Fred. Zwarts wrote:
>> Op 03.sep.2024 om 20:40 schreef olcott:
>>> On 9/3/2024 9:42 AM, joes wrote:
>>>> Am Mon, 02 Sep 2024 16:06:24 -0500 schrieb olcott:
>>>>> On 9/2/2024 12:52 PM, Fred. Zwarts wrote:
>>>>>> Op 02.sep.2024 om 18:38 schreef olcott:
>>>>>>> A halt decider is a Turing machine that computes the mapping from 
>>>>>>> its
>>>>>>> finite string input to the behavior that this finite string 
>>>>>>> specifies.
>>>>>>> If the finite string machine string machine description specifies 
>>>>>>> that
>>>>>>> it cannot possibly reach its own final halt state then this machine
>>>>>>> description specifies non-halting behavior.
>>>> Which DDD does not.
>>>
>>> DDD emulated by HHH cannot possibly reach
>>> its final halt state no matter what HHH does.
>>>
>>>>>>> A halt decider never ever computes the mapping for the computation
>>>>>>> that itself is contained within.
>>>> Then it is not total.
>>>
>>> Yes it is you are wrong.
>>>
>>>>>>> Unless there is a pathological relationship between the halt 
>>>>>>> decider H
>>>>>>> and its input D the direct execution of this input D will always 
>>>>>>> have
>>>>>>> identical behavior to D correctly simulated by simulating halt 
>>>>>>> decider
>>>>>>> H.
>>>> Which makes this pathological input a counterexample.
>>>
>>> Which makes the pathological input a counter-example
>>> to the false assumption that the direct execution of
>>> a machine always has the same behavior as the machine
>>> simulated by its pathological simulator.
>>>
>>>>>>> A correct emulation of DDD by HHH only requires that HHH emulate the
>>>>>>> instructions of DDD** including when DDD calls HHH in recursive
>>>>>>> emulation such that HHH emulates itself emulating DDD.
>>>>>> Indeed, it should simulate *itself* and not a hypothetical other HHH
>>>>>> with different behaviour.
>>>>> It is emulating the exact same freaking machine code that the x86utm
>>>>> operating system is emulating.
>>>
>>>> It is not simulating the abort because of a static variable. Why?
>>>>
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    OutputString("This code is unreachable by DDD emulated by HHH");
>>> }
>>>
>>>>>> If HHH includes code to see a 'special condition' and aborts and 
>>>>>> halts,
>>>>>> then it should also simulate the HHH that includes this same code and
>>>>> DDD has itself and the emulated HHH stuck in recursive emulation.
>>>
>>>> Your HHH incorrectly changes behaviour.
>>>>
>>>
>>> No you are wrong !!!
>>>
>>
>> Yes it does. HHH simulates only a few recursions, then it sees a 
>> 'special condition', stops the simulation, returns to DDD and DDD halts.
> 
> void DDD()
> {
>    HHH(DDD); // Fred lacks the software engineering skill to understand
>    OutputString("This code is unreachable from DDD emulated by HHH");
> }
> 
> 

Again, olcott language processing seems to be very poor.
Olcott does not even understand that I repeatedly agreed that HHH fails 
to reach that point.
HHH fails to reach the end of the halting program.
Olcott thinks that this failure is a proof for non-halting behaviour, 
but it only shows that the simulation was aborted too soon.
The finite string given to HHH describes a halting program, as proved by 
the direct execution, by the unmodified world class simulator and even 
by HHH1.
But HHH fails to reach the end of the program described by the finite 
string. No surprise, because HHH cannot possibly simulate itself 
correctly up to the end.